JAMB Mathematics Past Questions & Answers - Page 315

1,571.

A student measures a piece of rope and found that it was 1.26m long. If the actual length of the rope was 1.25m, what was the percentage error in the measurement?

A.

0.25%

B.

0.01%

C.

0.80%

D.

0.40%

Correct answer is C

Actual length of rope = 1.25m

Measured length of rope = 1.26m

error = (1.26 - 1.25)m - 0.01m

Percentage error = \(\frac{error}{\text{actual length}}\) x 100%

= \(\frac{0.01}{1.25}\) x 100%

= 0.8%

1,572.

Simplify (\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\)

A.

\(\frac{1}{5}\)

B.

\(\frac{1}{4}\)

C.

\(\frac{1}{36}\)

D.

\(\frac{1}{25}\)

Correct answer is C

(\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\)

Applying the rule of BODMAS, we have:

(\(\frac{3}{4}\) x \(\frac{4}{9}\) \(\div\)\(\frac{19}{2}\)) \(\div\)\(\frac{24}{19}\)


(\(\frac{1}{3}\) x \(\frac{2}{19}\)) \(\div\)\(\frac{24}{19}\)

\(\frac{1}{3}\) x \(\frac{2}{19}\) x \(\frac{19}{24}\)

= \(\frac{1}{36}\)

1,573.

Find r, if 6r7\(_8\) = 511\(_9\)

A.

3

B.

2

C.

6

D.

5

Correct answer is A

6r7\(_8\) = 511\(_9\)

6 x 8\(^2\) + r x 8\(^1\) + 7 x 8\(^0\) = 5 x 9\(^2\) + 1 x 9\(^1\) + 1 x 9\(^0\)

6 x 64 + 8r + 7 x 1 = 5 x 81 + 9 + 1 x 1

384 + 8r + 7 = 405 + 9 + 1

 

8r = 415 - 391

 8r = 24

r = \(\frac{24}{8}\)

= 3

1,574.

Find the derivative of \(\frac {\sin\theta}{\cos\theta}\)

A.

sec2 \(\theta\)

B.

tan \(\theta\) cosec \(\theta\)

C.

cosec \(\theta\)sec \(\theta\)

D.

cosec2\(\theta\)

Correct answer is A

\(\frac {\sin\theta}{\cos\theta}\)

\(\frac{\cos \theta {\frac{d(\sin \theta)}{d \theta}} - \sin \theta {\frac{d(\cos \theta)}{d \theta}}}{\cos^2 \theta}\)

\(\frac{\cos \theta. \cos \theta - \sin \theta (-\sin \theta)}{cos^2\theta}\)

\(\frac{cos^2\theta + \sin^2 \theta}{cos^2\theta}\)

Recall that sin2 \(\theta\) + cos2 \(\theta\) = 1

\(\frac{1}{\cos^2\theta}\) = sec2 \(\theta\)

1,575.

Find the probability that a number picked at random from the set(43, 44, 45, ..., 60) is a prime number.

A.

\(\frac{2}{3}\)

B.

\(\frac{1}{3}\)

C.

\(\frac{2}{9}\)

D.

\(\frac{7}{9}\)

Correct answer is C

Prime numbers = (43,47,53,59)

N = (43, 44, 45,..., 60)

The universal set contains 18 numbers.

The prime numbers between 43 and 60 are 4

Probability of picking a prime number = \(\frac{4}{18}\)

= \(\frac{2}{9}\)